Question

A function \( f : \mathbb{R} \to \mathbb{R} \) satisfies \[ f\!\left(\frac{x+y}{3}\right) = \frac{f(x) + f(y) + f(0)}{3} \quad \text{for all } x,y \in \mathbb{R}. \] If \( f'' \) is differentiable at \( x = 0 \), then \( f \) is:

• A. linear
• B. quadratic
• C. cubic
• D. biquadratic

Hint:

Mean-type functional equations: Symmetric averaging usually implies linearity. Try polynomial substitution and compare degrees.

Answer 1

The Correct Option is A

Concept: This is a functional equation involving averaging. Such symmetric mean-type functional equations typically restrict functions to low-degree polynomials. Step 1: Try polynomial assumption. Assume: \[ f(x) = ax^n \] Substitute into equation: \[ f\!\left(\frac{x+y}{3}\right) = a\left(\frac{x+y}{3}\right)^n \] Right side: \[ \frac{ax^n + ay^n + f(0)}{3} \] Step 2: Check degree possibilities.

• If \( n \ge 2 \), LHS produces mixed terms like \( xy \), which do not appear on RHS.
• So higher degree terms are not possible.

Hence polynomial must be degree \( \le 1 \). Step 3: Conclusion. Thus: \[ f(x) = mx + c \] i.e., a linear function.